By convention the nabla operator acts to the right. For this reason we must be careful. In your formula
$$(\vec{a} \times \nabla )\times \vec{b}$$
Since nabla acts to the right it acts on b and not a
The left acting nabla would act on a and not b
$$(\vec{a} \times \overset{\leftarrow}{\nabla} )\times \vec{b}$$
This is nonstandard notation, but we can write it as
$$-(\nabla \times \vec{a} ) \times \vec{b}=\vec{b} \times (\nabla \times \vec{a} )$$
which is your second formula (with a minus sign)
The sum of the left and right forms is the bidirectional form
$$(\vec{a} \times \overset{\leftrightarrow}{\nabla} )\times \vec{b}=(\vec{a} \times \nabla )\times \vec{b}+\vec{b} \times (\nabla \times \vec{a} )$$
This is useful when you get to integrals because you will have
$$\int_{\partial V} (\vec{a} \times n )\times \vec{b} \, \mathrm{d}S=\int_V ((\vec{a} \times \nabla )\times \vec{b}+\vec{b} \times (\nabla \times \vec{a} )) \, \mathrm{d}V$$
Where we change a surface integral into a volume integral.

The short version if any of that was confusing is that since nabla acts to the right it acts on a and not b in one formula and b and not a in the other, they are very different.

I don't know if that identity has a name. I brought it up to illustrate that
$$( {\vec a \times \nabla } ) \times \vec b \text{ and } ( {\nabla \times \vec a} ) \times \vec b$$
have different vectors being operated on while
$$( {\vec a \times \nabla } ) \times \vec b \text{ and } ( {\nabla \times \vec b} ) \times \vec a$$
have the same vector being operated on
These two forms arise in the gradient of dot product identity
$$\nabla_\vec{b} (\vec{a} \cdot \vec{b})=(\vec{a}\cdot \nabla)\vec{b}+\vec{a} \times (\nabla \times \vec{b})=\vec{a}(\nabla \cdot \vec{b})+(\vec{a} \times \nabla) \times \vec{b}$$
The nabla subscript b means that a is not operated on.

In your original question the key point was that one expression had a being operated on while the other had b being operated on. This is similar to in single variable calculus if we have
(D indicates the derivative)
D(uv)=uDv+vDu
uDv and vDu are not equal because
in uDv
v is being operated on while u is not
and
in vDu
u is being operated on while v is not